The vectors $\mathbf{a} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix}.$  There exist scalars $p,$ $q,$ and $r$ such that
\[\begin{pmatrix} 4 \\ 1 \\ -4 \end{pmatrix} = p \mathbf{a} + q \mathbf{b} + r (\mathbf{a} \times \mathbf{b}).\]Find $r.$
Answer: We can compute that $\mathbf{a} \times \mathbf{b} = \begin{pmatrix} 3 \\ 3 \\ 6 \end{pmatrix}.$  From the given equation,
\[(\mathbf{a} \times \mathbf{b}) \cdot \begin{pmatrix} 4 \\ 1 \\ -4 \end{pmatrix} = p ((\mathbf{a} \times \mathbf{b}) \cdot  \mathbf{a}) + q 
((\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b}) + r ((\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{a} \times \mathbf{b})).\]Since $\mathbf{a} \times \mathbf{b}$ is orthogonal to both $\mathbf{a}$ and $\mathbf{b},$ $(\mathbf{a} \times \mathbf{b}) \cdot  \mathbf{a} = (\mathbf{a} \times \mathbf{b}) \cdot  \mathbf{b} = 0,$ so this reduces to
\[-9 = 54r.\]Hence, $r = \boxed{-\frac{1}{6}}.$